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Proton Bombardment. A proton with mass \(1.67 \times10^{-27} \mathrm{kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) directiy toward a uranium nucleus 5.00 \(\mathrm{m}\) away. The proton is repelled by the uranium nucleus with a force of magnitude \(F=\alpha / x^{2}\) , where \(x\) is the separation between the two objects and \(\alpha=2.12 \times\) \(10^{-26} \mathrm{N} \cdot \mathrm{m}^{2} .\) Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10} \mathrm{m}\) from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the pro- ton get? (c) What is the speed of the proton when it is again 5.00 \(\mathrm{m}\) away from the uranium nucleus?

Step by step solution

01

Understand the Problem

This problem involves calculating the speed and closest distance of a proton as it moves towards and then away from a uranium nucleus under the effect of an inverse square law repelling force. We need to determine these elements using the laws of conservation of energy.

02

Calculate Initial Kinetic Energy

The initial kinetic energy (KE) of the proton can be calculated using the formula: \( KE = \frac{1}{2} m v^2 \), where \( m = 1.67 \times 10^{-27} \text{ kg} \) and \( v = 3.00 \times 10^{5} \text{ m/s} \). Substituting these values gives:\[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (3.00 \times 10^{5})^2 = 7.515 \times 10^{-17} \text{ J}\]

03

Calculate Potential Energy at a Position

Potential energy (PE) due to the force \( F = \frac{\alpha}{x^2} \) from a distance \( d \) to \( x \) can be calculated by integrating the force from \( d \) to \( x \). The potential energy at a distance \( x \) is:\[ PE = -\int_{\infty}^{x} \frac{\alpha}{x'^2} dx' = \frac{\alpha}{x}\]When \( x = 8.00 \times 10^{-10}\text{ m} \), \( PE = \frac{2.12 \times 10^{-26}}{8.00 \times 10^{-10}} = 2.65 \times 10^{-17} \text{ J}\).

04

Apply Conservation of Energy for Speed at Given Distance

Conservation of energy implies that total energy remains constant. Initially, total energy is only kinetic, while at distance \( x = 8.00 \times 10^{-10} \text{ m} \), it consists of both kinetic and potential energy. Set initial KE equal to the sum of KE and PE at \( x \):\[ KE_{initial} = KE + PE\]\[ 7.515 \times 10^{-17} = \frac{1}{2} m v^2 + 2.65 \times 10^{-17}\]Solving for \( v \), we get:\[ v = \sqrt{ \frac{2 (7.515 \times 10^{-17} - 2.65 \times 10^{-17})}{1.67 \times 10^{-27}} } = 2.50 \times 10^{5} \text{ m/s}\]

05

Determine Closest Approach

At the closest point of approach, all kinetic energy is converted to potential energy (the proton comes momentarily to rest). Equate initial kinetic energy to potential energy at closest distance \( x_c \):\[ 7.515 \times 10^{-17} = \frac{\alpha}{x_c}\]Solving for \( x_c \), we find:\[ x_c = \frac{\alpha}{7.515 \times 10^{-17}} = \frac{2.12 \times 10^{-26}}{7.515 \times 10^{-17}} = 2.82 \times 10^{-10} \, \text{m}\]

06

Speed at Original Distance after Repulsion

On returning to \( x = 5.00 \text{ m} \), potential energy is essentially zero because separation is large. By conservation of energy:\[ KE_{final} = KE_{initial}\]Speed at this point will be:\[ v = \sqrt{2 \times \frac{7.515 \times 10^{-17}}{1.67 \times 10^{-27}}} = 3.00 \times 10^{5} \text{ m/s}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It can be easily calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.
This formula shows that kinetic energy depends on both how much stuff the object has (mass) and how fast it’s moving (velocity squared).

• In our problem, the kinetic energy helps determine the proton's motion as it gets influenced by repulsive forces from the uranium nucleus.
• Initially, the proton has kinetic energy determined by its given speed of \( 3.00 \times 10^5 \text{ m/s} \).

When a proton approaches a force like the one from the uranium nucleus, its kinetic energy can be converted into potential energy or vice versa. This conversion is key to solving our exercise, as it enables us to apply the conservation of energy principle.

Potential Energy

Potential energy in our example is related to the position of the proton relative to the uranium nucleus. This kind of energy is stored due to its position and the force acting against it.
The potential energy in this context arises because of the repulsive force between the positively charged proton and the uranium nucleus.

• This force has a formula that follows the inverse square law: \( F = \frac{\alpha}{x^2} \) where \( x \) is the distance between the two particles.
• By integrating this force over distance, we obtain the potential energy expression: \( PE = \frac{\alpha}{x} \).

Potential energy is crucial when analyzing how close the proton gets to the nucleus before stopping and reversing its path, as it highlights points where kinetic energy is entirely converted to potential energy.

Proton Bombardment

In proton bombardment, a proton is fired towards another nucleus or particle. In this specific problem, we observe a proton directed at a uranium nucleus. The interaction involves electromagnetic forces that are governed by the properties of both particles.
When a proton nears a uranium nucleus, it experiences repulsion due to their like charges.

• The force involved is not constant; it changes with distance according to an inverse square pattern.
• By understanding this force, we can predict where and when the proton will stop moving closer—its closest approach—as it loses kinetic energy and gains potential energy.

This understanding allows us to calculate significant milestones in the proton's motion, such as its speed at certain distances and the point of closest approach.

Inverse Square Law

The inverse square law is a key principle for understanding the forces involved in proton bombardment. It states that the force between two charged particles is inversely proportional to the square of the distance between them. In simpler terms, as the distance increases, the force decreases rapidly.
For the problem at hand:

• The repulsive force between the proton and uranium nucleus decreases sharply as the proton moves away.
• Mathematically, the force is described with \( F = \frac{\alpha}{x^2} \); thus, doubling the distance reduces the force to a quarter.

This relationship is vital in solving for the potential energy and analyzing the proton's closest approach. By applying the inverse square law, we can understand how distance impacts the interactions between charged particles.